Question

Polynomial functions - Rational zero test: (see attachment below)
- in this answer how is the rational zero test being used to list all posible rational zeros?
- how are the rational zeros being found?

Answer 1

question 59 is "f(x) = x^3 + 3x^2 - x - 3"

Answer 2

and i need to use the rational zero test to list all possible zeros of f. then find the rational zeros

Answer 3

Rational zeros are found so that when you plug in x p(x)=0

Answer 4

but in the answer above - like idk what the heck they are doing...

Answer 5

where is the "f(x) = x^2(x+3)- (x+3) = (x+3) (x^2-1)" coming from?

Answer 6

like how do you solve for rational zeros? (different from possible rational zeros)

Answer 7

ok... so the rational root test says if you have a polynomial

and I'll use a cubic

\[P(x) = ax^3 + bx^2 + cx + d\]

find the factors of the constant d

so in your question find the factors of -3, call the factors p

then find the factors of the coefficient of the leading term ... so factors of a

so you need the factors of 1 call them q

the the rational root theorem says any rational root is

\[\frac{p}{q}\]

so the factors of -3 are p = 1 , -3 or -1 and 3

factors of 1 q = 1 and 1 or -1 and -1....

so loot at all the possible arrangements of p/q

so you basically get -1, 1, 3 and -3

so if you now use the factor theorem

P(-1) = P(1) = P(-3) = and P(-3)

any that equal zero are factors and hence you know the roots

Answer 8

where does (x + 3) come from

so using the rational root and factor theorem

you have x = -3 as a possible root

\[P(-3) =(-3)^3 + 3(-3)^2 -(-3) - 3\]

so P(-3) = 0 then x = -3 is a root so x + 3=0 is a factor

Answer 9

then if you look at the polynomial and use grouping in pairs

\[P(x) =( x^3 + 3x^2) - (x + 3) = x^2(x +3) - 1(x + 3)\]

Answer 10

why do they, after finding the possible rational zeros go ahead and write " "f(x) = x^2(x+3)- (x+3)"? how does that equal "(x+3) (x^2-1)"?

Answer 11

then you get the factorisation

\[P(x) = (x + 3)(x^2 -1) = (x +3)(x -1)(x + 1)\]

Answer 12

well the question is poorly structured... because when I 1sst saw it, I straight way when to grouping in pairs and then factoring...

so the rational root theorem wasn't needed....

but is you can find a possible rational root, and test it using substitution and show it's a zero, then the next step would be polynomial division or synthetic division...

I hope that makes sense

Answer 13

*and i understand now how they got the possible zeros +/- 1 and +/- 3. the steps after confuse me

Answer 14

well in a normal question, you could test all possible values... which you could have here

then come up with

P(-3) = 0, P(1) = 0 and P(-1) = 0

so then the linear factors are

P(x) = (x +3)(x-1)(x + 1)

and then distributed to show you got the original equation.,

Answer 15

Oh! ok and then you solve each component of P(x) = (x +3)(x-1)(x + 1) to 0 to get the final answer - the rational zeros of +/- 1, and -3

Answer 16

it's just another way to solve it - the answer they got in the textbook/attachment. how did they do that though- what does it all mean?

Answer 17

How to solve the general cubic explicitly for x ? I followed along once back in high school, but forget

Answer 18

I think the way the question and solution are shown is confusing...

the factoring defeats the goal of the question...

Answer 19

"f(x) = x^2(x+3)- (x+3)" equaling "(x+3) (x^2-1)" is very, very confusing..

Answer 20

what is shown there is called grouping in pairs...

so you have a common factor of (x +3) whats left when its removed is X^2 - 1

so then the factors are

(x +3)(x^2 - 1)

Answer 21

and the quadrtic factor is just the difference of 2 squares

Answer 22

sorry here's another question - asking to the same thing except with a different function- f(x) = 2x^4-17x^3+35x^2+9x -45

Answer 23

how is the synthetic division used?

Answer 24

This is interesting. have fun, got to go

Answer 25

http://www.sosmath.com/algebra/factor/fac11/fac11.html

Answer 26

so start with rational roots

\[45 = \pm 1 \pm 3 \pm 5 \pm 9 \pm 15~~~and ~~~2 = \pm1, \pm 2\]

so looking at it and some quick mental arithmetic try x = -1

and so find P(-1) if it equals zero, you have a factor and can use synthetic division or polynomial division.

Answer 27

so if P(-1) is a zero, then x + 1 is a factor

Answer 28

so to find p(-1) = 0, you pug in -1 for x into the original function right?

Answer 29

* plug

Answer 30

and would you keep on doing that for all factors of 45, and 2?

Answer 31

correct...

Answer 32

and for which ever numbers that equal 0 then you would use synthetic division to verify then?

Answer 33

that seems so time consuming, and bad for tests, is there a short cut, especially if you cant use calculators on a quiz?

Answer 34

could... but given you have 2x^4.... somewhere along the way you will have a linear factor (2x + ?)

so when I know 1 linear factor I'd probably do a division of some sort... or try anothe rational zero, maybe P(3) or P(-3)

Answer 35

so i have 2 options then to check if a factor is zero: plugging it back into the equation OR synthetic division, correct?

Answer 36

that's it... if you can shown a 2nd rational root... then you can do the division...

knowing when you get the cubic you can divide it by the next factor...

Answer 37

I just did the 1st stage of a polynomial division and I get

\[P(x) = (x +1)(2x^3 - 19x^2 + 54x - 45)\]

you still have -45 so see what happens with P(3) and P(-3)

if either of those work you will be able to divide the cubic and get a quadratic

Answer 38

I just checked and P(3) = 0 so x - 3 is a factor...

so you will get down to a quadratic with the coefficient of the leading term being 2

and it can be factored..

Hope it all made sense